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Standard XII

Physics

Multiplication with Vectors

Two tall buil...

Question

Two tall buildings are $80 m$ apart. The velocity with which a ball should be thrown horizontally from a window $95 m$ above the ground in one building so that it will enter a window $15 m$ above the ground in the second building is : ( $g = 10 m / s^{2}$ )

15 m / s

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5 m / s

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10 m / s

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20 m / s

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Solution

The correct option is D $20 m / s$
Range $= 80 m$
Height $= 80 m$
$u_{v} = 0$
$S = - 80 m$
$a = - 10 m / s^{2}$
$- 80 = 0 - \frac{1}{2} \times 10 \times t^{2} \Rightarrow t = 4 s e c$

Range $= u \times t$
$\Rightarrow u = \frac{R a n g e}{t} = \frac{80}{4} = 20 m / s$

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Two tall buildings are 80 m apart. The velocity with which a ball should be thrown horizontally from a window 95 m above the ground in one building so that it will enter a window 15 m above the ground in the second building is : (g=10 m/s2)

The correct option is D 20 m/sRange =80 mHeight =80 muv=0S=−80 m a=−10 m/s2−80=0−12×10×t2⇒t=4 secRange =u×t⇒u=Ranget=804=20 m/s

Two tall buildings are $80 m$ apart. The velocity with which a ball should be thrown horizontally from a window $95 m$ above the ground in one building so that it will enter a window $15 m$ above the ground in the second building is : ( $g = 10 m / s^{2}$ )

The correct option is D $20 m / s$
Range $= 80 m$
Height $= 80 m$
$u_{v} = 0$
$S = - 80 m$
$a = - 10 m / s^{2}$
$- 80 = 0 - \frac{1}{2} \times 10 \times t^{2} \Rightarrow t = 4 s e c$

Range $= u \times t$
$\Rightarrow u = \frac{R a n g e}{t} = \frac{80}{4} = 20 m / s$